Question

solve for X = 2= square root (2x−5)−square root(x−2)


Answer 1

\[Solve for X: 2=\sqrt{2x-5}-\sqrt{x-2}\]

Answer 2

\[\huge{2=\sqrt{2x-5}-\sqrt{x-2}}\]
\[\huge{2^2=(\sqrt{2x-5}-\sqrt{x-2})^2}\]
\[\huge{4=2x-5+x-2-2\sqrt{2x-5}\sqrt{x-2}}\]
\[\huge{4=3x-7-2\sqrt{(2x-5)(x-2)}}\]
\[\huge{11=3x-2\sqrt{2x^2-4x-5x+10}}\]
\[\huge{11=3x-2\sqrt{2x^2-9x+10}}\]

Answer 3

can u do further ?

Answer 4

how did you get step #3?

Answer 5

wouldnt it be 2x-5-x-2? thank you for the answer btw. I'm just trying to understand the steps :)

Answer 6

What he did at step 3 is basically do binomial multiplication.

Answer 7

oh i see. how would i then solve for x after the last step?

Answer 8

\[\huge(-\frac{11 - 3x}{2})^{2} = (\sqrt{2x^{2} - 9x + 10})^{2}\]

Answer 9

Can you finish this?

Answer 10

I'm sorry i don't quite understand haha. I'm completely stumped

Answer 11

\[\huge\frac{9x^{2} - 66x + 121}{4} = 2x^{2} - 9x + 10\]@mathslover Did I do it right?

Answer 12

@mathslover look ^ did I do it right?

Answer 13

u did right

Answer 14

Alright. Now from what I did above:
\[9x^{2} - 66x + 121 = 4(2x^{2} - 9x + 10)\]\[9x^{2} - 66x + 121 = 8x^{2} - 36x + 40\]\[x^{2}- 30x + 81\]Can you solve form here?

Answer 15

That last line should equal 0.

Answer 16

got it (x-27)(x-3)=0. so 27 and 3

Answer 17

thanks again calcmathlete

Answer 18

Yeah. Now plug it back in for x and see if both solutions work.

Answer 19

and @mathslover

Answer 20

only 27 works

Answer 21

Wait, before you finish, be sure to plug 27 and 3 back in to check for extraneous solutions.

Answer 22

Yeah. That means x = 27 and ≠ 3

Answer 23

Got it. thanks again!